I would like to "relate" two arithmetic averages and then use them in Poisson. I currently already use a method, but i'm sure we can do better and that there is a better way. I currently sum the two averages, then divide by two, then use the result in Poisson.

In the championship matches played so far, Team_A and Team_B have never clashed against each other, but so far the average number of goals scored by Team_A is 1.40; Instead the average number of goals conceded by Team_B is 1.80. I would like to calculate how many probabilità there are that Team_A scores 2 goals against Team_B in the next match when the two teams will clash against each other. The two averages are independent, because there is no correlation between the two datasets.

An important observation is that: the goals scored by Team_A against Team_B will be the same as those that Team_B will concede from Team_A.

Is there any better way that can replace addition and division by 2? (my Union) and find a value to use in Poisson?

TEAM_A__List_Goals = 2, 1, 3, 0, 1
TEAM_A__Total_Goals_Scored = 7
TEAM_A__Number_Match_Played = 5
TEAM_A__Average_Scored = 1.40

TEAM_B__List_Goals = 1, 2, 2, 3, 1
TEAM_B__Total_Goals_Conceded = 9
TEAM_B__Number_Match_Played = 5
TEAM_B__Average_Conceded = 1.8

Union = (TEAM_A__Goals_Scored + TEAM_B__Goals_Conceded) / 2

two_goals = ((Union ** 2) * 2.7182818284 ** (-Union)) / 2 * 100

Thank you all!

  • $\begingroup$ I assume your RV are poisson distributed with $\mathbb{E}(A)=1.4$ and $\mathbb{E}(B)=1.8$. You can now express the probability that A scores atleast two goals as a conditional probability $\mathbb{P}(A \geq 2 | A=B)$. $\endgroup$ Nov 15, 2023 at 11:17
  • $\begingroup$ @Keine_Maschine In this case, what would the final result (parameter to be inserted in Poisson) be? $\endgroup$ Nov 15, 2023 at 11:20
  • $\begingroup$ You can calculate the distribution $\mathbb{P}(X=0), \mathbb{P}(X=1), ...$ with $\mathbb{P}(X=i)=\mathbb{P}(A=i|A=B)$. I'm not sure wheter $X$ is still poisson distributed, but that doesn't really matter. If it is you can estimate the parameter for example by regression. $\endgroup$ Nov 15, 2023 at 11:30
  • $\begingroup$ @Keine_Maschine Is the formula you showed me that of Conditional Probability? Can you show me the numerical result of your formula? (and how to get it). Please :) $\endgroup$ Nov 15, 2023 at 12:19
  • $\begingroup$ You can google the formula for conditional probability, then you can calculate the distribution of $X$. Example: $\mathbb{P}(X=0)=\mathbb{P}(A=0|A=B)=\frac{\mathbb{P}(A=B=0)}{\mathbb{P}(A=B)}=\frac{\mathbb{P}(A=0) \cdot \mathbb{P}(B=0) }{\mathbb{P}(A=B)}$. $\mathbb{P}(A=0)$ and $\mathbb{P}(B=0)$ are clear, $\mathbb{P}(A=B)$ is open to calculate. I think you can do this on your own! :) $\endgroup$ Nov 15, 2023 at 14:49


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